Problem: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $23.8$ years; the standard deviation is $5.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living less than $18.3$ years.
Answer: $23.8$ $18.3$ $29.3$ $12.8$ $34.8$ $7.3$ $40.3$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $23.8$ years. We know the standard deviation is $5.5$ years, so one standard deviation below the mean is $18.3$ years and one standard deviation above the mean is $29.3$ years. Two standard deviations below the mean is $12.8$ years and two standard deviations above the mean is $34.8$ years. Three standard deviations below the mean is $7.3$ years and three standard deviations above the mean is $40.3$ years. We are interested in the probability of a zebra living less than $18.3$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the zebras will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the zebras will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $18.3$ years and the other half $({16\%})$ will live longer than $29.3$ years. The probability of a particular zebra living less than $18.3$ years is ${16\%}$.